Bourbaki

Proposition 3

Tag 05PY, permanent. Cite it as [Bourbaki, Tag 05PY].

Algebra, chapter II, § 4. Relations between tensor products and homomorphism modules

Let CC be a commutative ring, E,FE, F two finitely generated projective CC-modules* and u:E→Fu : E \to F and v:F→Ev : F \to E two linear mappings; then*

Tr(v∘u)=Tr(u∘v). \mathrm{Tr}(v \circ u) = \mathrm{Tr}(u \circ v).

The two mappings (u,v)↦Tr(u∘v),(u,v)↦Tr(v∘u)(u, v) \mapsto \mathrm{Tr}(u \circ v), (u, v) \mapsto \mathrm{Tr}(v \circ u) of

HomC(E,F)×HomC(F,E) \mathrm{Hom}_C(E, F) \times \mathrm{Hom}_C(F, E)

into CC are CC*-bilinear*; it therefore suffices to verify (20) when uu is of the form x↦⟨x,a∗⟩bx \mapsto \langle x, a^* \rangle b and vv of the form y↦⟨y,b∗⟩ay \mapsto \langle y, b^* \rangle a, with a∈E,a∗∈E∗,b∈F,b∗∈F∗a \in E, a^* \in E^*, b \in F, b^* \in F^*. But then v∘uv \circ u is the mapping x↦⟨x,a∗⟩⟨b,b∗⟩ax \mapsto \langle x, a^* \rangle \langle b, b^* \rangle a and u∘vu \circ v the mapping y↦⟨y,b∗⟩⟨a,a∗⟩by \mapsto \langle y, b^* \rangle \langle a, a^* \rangle b. Formula (17) shows that the values of the two sides of (20) are equal to ⟨a,a∗⟩⟨b,b∗⟩\langle a, a^* \rangle \langle b, b^* \rangle.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

Nothing.

Transcribed from the printed edition, 1998, Springer. Not translated.